Using the Wronskian to Solve Differential Equations with Non-Constant Solutions

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Discussion Overview

The discussion revolves around the use of the Wronskian to solve a specific differential equation of the form y'' + ay' + by = 0, where a and b are real constants. Participants explore how to demonstrate the existence of constants d₁ and d₂ such that certain linear combinations of solutions v₁ and v₂ yield any solution f(x) of the differential equation. The focus is on the relationships derived from the Wronskian and the implications of its non-zero value.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that since the quotient v₁/v₂ is not constant, the Wronskian must be non-zero.
  • Participants derive expressions for d₁ and d₂ based on equations (1) and (2) but question whether these derivations align with the requirement to use Wronskian identities.
  • There is a discussion about the solvability of the system of equations for d₁ and d₂, particularly when the Wronskian is non-zero.
  • One participant suggests that the existence of d₁ and d₂ is contingent upon the non-zero denominator in their expressions, implying that solutions would be undefined otherwise.
  • Another participant seeks clarification on whether the two relations (1) and (2) are given or need to be proven, leading to a consensus that they have been proven through the discussion.
  • The general solution form y = d₁v₁(x) + d₂v₂(x) is mentioned, with a query on how to establish that constants A and B correspond to d₁ and d₂.

Areas of Agreement / Disagreement

Participants generally agree on the necessity of the Wronskian being non-zero for the existence of solutions d₁ and d₂. However, there is some uncertainty regarding the derivation of the general solution and whether the relations (1) and (2) were initially given or need to be proven.

Contextual Notes

The discussion includes various mathematical manipulations and assumptions regarding the properties of the Wronskian and the independence of solutions, which may not be fully resolved or universally accepted among participants.

Who May Find This Useful

Readers interested in differential equations, particularly in the context of linear combinations of solutions and the application of the Wronskian in proving solution existence, may find this discussion beneficial.

mathmari
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Hey! :)
I need some help at the following exercise:
Let v_{1}, v_{2} solutions of the differential equation y''+ay'+by=0 (where a and b real constants)so that \frac{v_{1}}{v_{2}} is not constant.If y=f(x) any solution of the differential equation ,use the identities of the Wronskian to show that there are constants d_{1}, d_{2} so that:d_{1}v_{1}(0)+d_{2}v_{2}(0)=f(0) (1), d_{1}v_{1}'(0) +d_{2}v_{2}'(0)=f'(0) (2) and that each solution of the differential equation has the form:y=d_{1}v_{1}(x)+d_{2}v_{2}(x).

Since \frac{v_{1}}{v_{2}} is not constant, the Wronskian is not equal to zero, right? But how can I continue to show the relations (1) and (2)??
 
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mathmari said:
Hey! :)
I need some help at the following exercise:
Let v_{1}, v_{2} solutions of the differential equation y''+ay'+by=0 (where a and b real constants)so that \frac{v_{1}}{v_{2}} is not constant.If y=f(x) any solution of the differential equation ,use the identities of the Wronskian to show that there are constants d_{1}, d_{2} so that:d_{1}v_{1}(0)+d_{2}v_{2}(0)=f(0) (1), d_{1}v_{1}'(0) +d_{2}v_{2}'(0)=f'(0) (2) and that each solution of the differential equation has the form:y=d_{1}v_{1}(x)+d_{2}v_{2}(x).

Since \frac{v_{1}}{v_{2}} is not constant, the Wronskian is not equal to zero, right? But how can I continue to show the relations (1) and (2)??

Hi! :)

What do you get if you solve equations (1) and (2) for $d_1$ and $d_2$?
 
I like Serena said:
Hi! :)

What do you get if you solve equations (1) and (2) for $d_1$ and $d_2$?

Solving for $d_2$ I get, $d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$.

And $d_1=\frac{f(0)-d_2v_2(0)}{v_1(0)}$, where $d_2$ is the above relation.

But am I not asked to show the equations (1) and (2) by using the identities of Wronskian??
 
mathmari said:
Solving for $d_2$ I get, $d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$.

And $d_1=\frac{f(0)-d_2v_2(0)}{v_1(0)}$, where $d_2$ is the above relation.

But am I not asked to show the equations (1) and (2) by using the identities of Wronskian??

Just a sec...
Suppose you multiply both the numerator and the denominator by $v_1(0)$, what do you get?
In other words, exactly when is the system of those 2 equations solvable?
 
I like Serena said:
Just a sec...
Suppose you multiply both the numerator and the denominator by $v_1(0)$, what do you get?
In other words, exactly when is the system of those 2 equations solvable?

$d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$

$v_1(0)v'_2(0)-v_2(0)v'_1(0)!=0 $ ,that we have from the Wronskian.right?
 
mathmari said:
$d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$

That should be:
$$d_2=\frac{f'(0)v_1(0)-f(0)v'_1(0)}{v_1(0)v'_2(0)-v'_1(0)v_2(0)}
= \frac{f'(0)v_1(0)-f(0)v'_1(0)}{W(v_1, v_2)(0)}$$
$v_1(0)v'_2(0)-v_2(0)v'_1(0)!=0 $,that we have from the Wronskian.right?

Yes.
One of the identities of the Wronskian is, that if the two functions are independent (their quotient is well-defined and is not constant), then the Wronskian is different from zero.
Therefore the equation has a guaranteed solution! ;)
 
I like Serena said:
That should be:
$$d_2=\frac{f'(0)v_1(0)-f(0)v'_1(0)}{v_1(0)v'_2(0)-v'_1(0)v_2(0)}
= \frac{f'(0)v_1(0)-f(0)v'_1(0)}{W(v_1, v_2)(0)}$$

Yes.
One of the identities of the Wronskian is, that if the two functions are independent (their quotient is well-defined and is not constant), then the Wronskian is different from zero.
Therefore the equation has a guaranteed solution! ;)

And what does this mean? Have we shown now that there are $d_1$ and $d_2$ that satisfy the two relations? Do these two constants exist because the denominator is different from zero?
 
mathmari said:
And what does this mean? Have we shown now that there are $d_1$ and $d_2$ that satisfy the two relations? Do these two constants exist because the denominator is different from zero?

Together with the solution for $d_1$, which is similar, we have shown that there are constants $d_1$, $d_2$ so that: $d_1v_1(0)+d_2v_2(0)=f(0)$ (1), $d_1v′_1(0)+d_2v′_2(0)=f′(0)$ (2).

Indeed, these solutions only exist because the denominator for both $d_1$ and $d_2$ are different from zero. Otherwise they would be undefined. That is, a solution would not exist.

Btw, how do you feel about picking an avatar picture? :o
 
I like Serena said:
Together with the solution for $d_1$, which is similar, we have shown that there are constants $d_1$, $d_2$ so that: $d_1v_1(0)+d_2v_2(0)=f(0)$ (1), $d_1v′_1(0)+d_2v′_2(0)=f′(0)$ (2).

Indeed, these solutions only exist because the denominator for both $d_1$ and $d_2$ is different from zero. Otherwise they would be undefined. That is, a solution would not exist.

These two relations are given, or do I have to prove them?
 
  • #10
mathmari said:
These two relations are given, or do I have to prove them?

These two relations are not given.
However, we have just proven that they hold.
 
  • #11
I like Serena said:
These two relations are not given.
However, we have just proven that they hold.

A ok. :D

How can I show that the general solution is $y=d_1v_1(x)+d_2v_2(x)$?
 
  • #12
Since $v_1$ and $v_2$ are solutions of $y$, the general solution is $y=Av_1+Bv_2$. But how could we show that $A=d_1$ and $B=d_2$?
 

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